Bilinears in adjoint representation

Question

Below are two statements from my notes and I am trying to verify them explicitly. In both cases the fields are assumed to transform under the fundamental representation of $O(N)$ -

--'The kinetic term for a Dirac spinor is invariant under the symmetry group $U(N) \otimes U(N)$'.

I considered the case of a Weyl spinor first. This has a kinetic term $i \bar \phi_R \gamma^{\mu} \partial_{\mu} \phi_R$ and if $\phi_R \rightarrow U \phi_R$ then $i \bar \phi_R \gamma^{\mu} \partial_{\mu} \phi_R \rightarrow i\phi^{\dagger}_R U^{\dagger} \gamma_0 \gamma^{\mu} \partial_{\mu} U \phi $. Because $U$ and the gamma matrices act on different spaces, can I just shift the $U$ to the $U^{\dagger}$ and then using $UU^{\dagger}=1$ get the result? The $U(N) \otimes U(N)$ for the Dirac spinors comes about from decomposing a Dirac spinor into its left and right handed components each of which transforms under a 'left handed fundamental representation' or 'right handed fundamental representation' so could write the symmetry group as $U_L(N) \otimes U_R(N)$ (correct if mistaken).

--'If $T_a$ are the generators of $O(N)$, the bilinears $\phi^T T^a \phi$ transform according to the adjoint representation.'

I'm just wondering do generators always transform in the adjoint representation themselves? I read on here in another thread that the adjoint representation can be thought of as the representation anchored at the identity so if anyone could shed some light on this statement that'd be great.

Answer 1

Yes, the first part of your question is appreciated and answered soundly. The fermion kinetic term splits into two independent parts involving left and right Weyl spinors respectively, so each is independent under a separate U(N) as your wrote down.

The second question is a matter of language. A generator is a matrix with one adjoint index, a here, ranging over the dimension of the Lie algebra, so "universal" as per your question; and two indices each corresponding to its representation, e.g. i,j, the matrix indices, ranging over the dimension of that particular representation. It is an operator of the representation, acting on vectors of it, such as your φ. If your φ is in the fundamental, for instance, T will act on it as $\phi_i\mapsto T^a_{ij} \phi_j$, all in the fundamental. Dotting this with another vector, φ will yield a scalar in the fundamental, your expression, with a loose index a of the adjoint, so, then a vector in the adjoint. Always in the adjoint, regardless of what irrep φ you started with, as long as you used the suitable representation matrices for that generator.

To rotate this vector under O(N), you'll have to act on it with the Lie algebra structure constants, which are the operators T in the adjoint, so, analogously, $\phi^T T^a \phi\mapsto i f_{abc}~\phi^T T^c \phi$. For instance, for O(3) the generators are the familiar vector spin matrices $i\epsilon_{abc}$s.

Answer 2

(I) Assuming there are $N$ distinct fermions in your Lagrangian (e.g. quarks in the standard model), the kinetic term will have the $U(N)\times U(N)$ symmetry. This symmetry is broken by the mass term, however, which couples fermions with different handedness.

(II) If $V$ is the fundamental representation of a Lie group (with dual/conjugate representation $V^*$), generators can be identified as a subspace of $V\otimes V^*$. If $g$ is an element of the Lie group, then $g$ 'acts on' $V\otimes V^*$ as follows: $g\cdot (v\otimes \omega)=(U(g)v)\otimes (U(g)' \omega)=U(g)v\otimes \omega U(g^{-1})$. (Often it's possible to choose a unitary representation, so that $U(g^{-1})=U(g)^\dagger$).

[Note that the tensor representation $V\otimes V^*$ is reducible, and only contains the adjoint representation as a subrepresentation].

Now it turns out that a large class of Lie groups can be expressed locally as the exponential of some element in the tangent space near the identity: i.e. $$ U(g)=\exp(-iA_a(g)\tau^a), $$ where $A_a(g)$ are some $g$ dependent coefficients (local coordinates for the Lie group), and $\tau^a$ are the generators. The exponential is interpreted as a power series, when sensible (when $V$ is infinite dimensional, as in quantum mechanics, it's typical to specify that the spectrum of $A_a\tau^a$ is bounded from below, so that analytic continuation can be used).

These local coordinates for the Lie group imply that the group action defined above induces an action of the tangent space near the identity (i.e. the Lie algebra associated with the group): $\mathcal A \cdot v\otimes \omega=[\mathcal A, M_{v\otimes\omega}]$, where $M_{v\otimes\omega}$ is the 'matrix' associated with $v\otimes \omega$.

('Trivial' (in the common sense) addendum: the reason higher order terms in the power series expansion of $U(g)$ do not give alternative adjoint representations of the Lie algebra is essentially a kind of 'memory effect', where the linear term interferes with higher order components).